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Yogita Ingle 6 years ago
When two different coins are tossed randomly, the sample space is given by
S = {HH, HT, TH, TT}
Therefore, n(S) = 4.
Getting two heads:
Let E1 = event of getting 2 heads. Then,
E1 = {HH} and, therefore, n(E1) = 1.
Therefore, P(getting 2 heads) = P(E1) = n(E1)/n(S) = 1/4.
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Yogita Ingle 6 years ago
Numbers we will use
=> 2930 - 7 = 2923
=> 3250 - 11 = 3239
Thus by Prime Factorisation Method
=> 2923 = 37 x 79
=> 3239 = 41 x 79
Thus HCF(2923,3239) = 79
Thus, the greatest number that will divide 2930 and 3250 to leave 7 and 11 as remainder.
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Aviral Goyal 6 years ago
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